Problem: Find $a$ if $a$ and $b$ are integers such that $x^2 - x - 1$ is a factor of $ax^{17} + bx^{16} + 1$.
Answer: If $x^2-x-1$ is a factor of $ax^{17}+bx^{16}+1,$ then both the roots of $x^2-x-1$ must also be roots of $ax^{17}+bx^{16}+1.$ Let $s$ and $t$ be the roots of $x^2-x-1.$ Then we must have \[as^{17} + bs^{16} + 1 = at^{17} + bt^{16} + 1 = 0.\]Since $s$ is a root of $s^2-s-1=0,$ we have $s^2=s+1.$ This equation lets us express higher powers of $s$ in the form $Ms+N,$ for constants $M$ and $N.$ We have \[\begin{aligned} s^3 &= s^2 \cdot s = (s+1)s = s^2+s=(s+1)+s=2s+1, \\ s^4 &= s^3 \cdot s = (2s+1)s = 2s^2 + s = 2(s+1) + s = 3s+2, \\ s^5 &= s^4 \cdot s =(3s+2)s = 3s^2+2s=3(s+1)+2s=5s+3, \end{aligned}\]and so on. Seeing a pattern, we guess that \[s^n = F_ns + F_{n-1},\]where $\{F_n\}$ are the Fibonacci numbers (with $F_1 = F_2 = 1,$ and $F_n = F_{n-1} + F_{n-2}$ for $n \ge 3$). We can prove this formula with induction (see below). This means that \[s^{16} = F_{16}s + F_{15} = 987s + 610 \; \text{ and } \; s^{17} = F_{17}s + F_{16} = 1597s + 987.\]Thus, \[as^{17} + bs^{16} + 1 = (1597a+987b)s + (987a+610b) + 1,\]so it must be the case that $1597a + 987b = 0$ and $987a + 610b =- 1.$ This system has solutions $a = \boxed{987}$ and $b = -1597.$

Proof of formula: We already did the base cases of the induction. If $s^n = F_ns + F_{n-1}$ for some value of $n,$ then \[\begin{aligned} s^{n+1} = s^n \cdot s &= (F_ns + F_{n-1}) \cdot s \\ &= F_ns^2 + F_{n-1}s\\ & = F_n(s+1) + F_{n-1}s\\ & = (F_n+F_{n-1})s + F_n = F_{n+1}s + F_n. \end{aligned}\]This completes the inductive step. $\square$